A、哪天返回
小明被不明势力劫持。后被扔到x星站再无问津。小明得知每天都有飞船飞往地球,但需要108元的船票,而他却身无分文。
他决定在x星战打工。好心的老板答应包食宿,第1天给他1元钱。
并且,以后的每一天都比前一天多2元钱,直到他有足够的钱买票。
请计算一下,小明在第几天就能凑够108元,返回地球。
题解:
package action; public class demo { public static void main(String[] args) { int an = 1; int sn = 0; int n=0; for (n = 0; sn < 108; n++) { sn += an; an += 2; } System.out.println(n); } }
B、猴子分香蕉
5只猴子是好朋友,在海边的椰子树上睡着了。这期间,有商船把一大堆香蕉忘记在沙滩上离去。
第1只猴子醒来,把香蕉均分成5堆,还剩下1个,就吃掉并把自己的一份藏起来继续睡觉。
第2只猴子醒来,重新把香蕉均分成5堆,还剩下2个,就吃掉并把自己的一份藏起来继续睡觉。
第3只猴子醒来,重新把香蕉均分成5堆,还剩下3个,就吃掉并把自己的一份藏起来继续睡觉。
第4只猴子醒来,重新把香蕉均分成5堆,还剩下4个,就吃掉并把自己的一份藏起来继续睡觉。
第5只猴子醒来,重新把香蕉均分成5堆,哈哈,正好不剩!
请计算一开始最少有多少个香蕉。
题解:
package action; public class demo { public static void main(String[] args) { for (int n = 20;; n++) { double y = n; // 桃子的个数 for (int x = 1; x <= 4; x++) { y = y - x - (y - x) / 5; } if (y % 5 == 0) { System.out.println(n); break; } } } }
C、字母阵列
仔细寻找,会发现:在下面的8x8的方阵中,隐藏着字母序列:“LANQIAO”。
SLANQIAO
ZOEXCCGB
MOAYWKHI
BCCIPLJQ
SLANQIAO
RSFWFNYA
XIFZVWAL
COAIQNAL
我们约定: 序列可以水平,垂直,或者是斜向;
并且走向不限(实际上就是有一共8种方向)。
上图中一共有4个满足要求的串。
下面有一个更大的(100x100)的字母方阵。
你能算出其中隐藏了多少个“LANQIAO”吗?
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题解:
package action; import java.util.Scanner; public class demo { static int n = 100; // 长度 static char[] ch = { 'L', 'A', 'N', 'Q', 'I', 'A', 'O' }; // 寻找数据 static char[][] data = new char[n][n]; // 存储字母方阵 // 从正北方向绕顺时针旋转一圈的八个方向对应的增减值 static int[] dx = { 0, 1, 1, 1, 0, -1, -1, -1 }; static int[] dy = { 1, 1, 0, -1, -1, -1, 0, 1 }; static int count = 0; // 计数 public static void main(String[] args) { Scanner sc = new Scanner(System.in); // 插入数据 for (int i = 0; i < data.length; i++) { data[i] = sc.nextLine().toCharArray(); } // 循环 for (int i = 0; i < data.length; i++) { for (int j = 0; j < data[i].length; j++) { // 找第一个字符L if (data[i][j] == ch[0]) { sheck(i, j); // 查找方法 } } } System.out.println(count); // 打印 sc.close(); } public static void sheck(int i, int j) { // 八个方向 for (int d = 0; d < 8; d++) { int x = i + dx[d]; int y = j + dy[d]; int num = 1; // 寻找字符数组下标 while (x < n && x >= 0 && y < n && y >= 0 && data[x][y] == ch[num]) { // 当测试到最后一个字母以后,计数+1,并跳出while循环,进行下一个方向的寻找 if (num == 6) { count++; break; } num++; // 下一个 // 对应坐标加上相应的dx,dy x += dx[d]; y += dy[d]; } } } }
D、第几个幸运数
到x星球旅行的游客都被发给一个整数,作为游客编号。x星的国王有个怪癖,他只喜欢数字3,5和7。
国王规定,游客的编号如果只含有因子:3,5,7,就可以获得一份奖品。
我们来看前10个幸运数字是:3 5 7 9 15 21 25 27 35 45
因而第11个幸运数字是:49
小明领到了一个幸运数字 59084709587505,他去领奖的时候,人家要求他准确地说出这是第几个幸运数字,否则领不到奖品。
请你帮小明计算一下,59084709587505是第几个幸运数字。
题解:
package action; public class demo { public static void main(String[] args) { long y = 59084709587505l; int count = 0; for (int i = 0; Math.pow(3, i) < y; i++) { for (int j = 0; Math.pow(5, j) < y; j++) { for (int k = 0; Math.pow(7, k) < y; k++) { if (Math.pow(3, i) * Math.pow(5, j) * Math.pow(7, k) < y) { count++; System.out.println(count + " " + Math.pow(3, i) * Math.pow(5, j) * Math.pow(7, k)); } } } } System.out.println(count); } }
E、书号验证
2004年起,国际ISBN中心出版了《13位国际标准书号指南》。
原有10位书号前加978作为商品分类标识;校验规则也改变。
校验位的加权算法与10位ISBN的算法不同,具体算法是:
用1分别乘ISBN的前12位中的奇数位(从左边开始数起),用3乘以偶数位,
乘积之和以10为模,10与模值的差值再 对10取模(即取个位的数字)即可得到校验位的值,
其值范围应该为0~9。
下面的程序实现了该算法,请仔细阅读源码,填写缺失的部分。
public class A { static boolean f(String s){ int k=1; int sum = 0; for(int i=0; i<s.length(); i++){ char c = s.charAt(i); if(c==’-’ || c==’ ') continue; sum += ______________________________; //填空 k++; if(k>12) break; } return s.charAt(s.length()-1)-‘0’ == (10-sum % 10)%10; } public static void main(String[] args){ System.out.println(f(“978-7-301-04815-3”)); System.out.println(f(“978-7-115-38821-6”)); } }
题解:
package action; public class demo { static boolean f(String s) { int k = 1; // 位数 int sum = 0; // 乘积之和 for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); // 获取每个字符 if (c == '-' || c == ' ') continue; // 跳转杠和空格 // (k % 2 == 0?3:1) : 判断是否是奇数还是偶数,并且得到对应的乘值 sum += (c - '0') * (k % 2 == 0 ? 3 : 1); // 填空位置 k++; // 每次循环位数+1 if (k > 12) break; // 超出12位直接跳出 } // 乘积之和以10为模,10与模值的差值再对10取模(即取个位的数字)即可 return s.charAt(s.length() - 1) - '0' == (10 - sum % 10) % 10; } public static void main(String[] args) { System.out.println(f("978-7-301-04815-3")); System.out.println(f("978-7-115-38821-6")); } }